A particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When particle is very far from the earth, its speed would be

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Aicharya · Accepted Answer

By conservation of mechanical energy, $\begin{aligned} & \mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}} \ & \Rightarrow \frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}ight)^2-\frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{E}}}=\frac{1}{2} \mathrm{mv}^2+0 \ & \Rightarrow 4 \mathrm{v}_2^2-\frac{2 \mathrm{GM}}{\mathrm{R}_{\mathrm{E}}}=\mathrm{v}^2 \ & \Rightarrow \mathrm{~V}^2=4 \mathrm{v}_{\mathrm{e}}^2-\mathrm{v}_{\mathrm{e}}^2 \quad\left[\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}_{\mathrm{E}}}}ight] \ & 	herefore \mathrm{v}=\sqrt{3} \mathrm{v}_{\mathrm{e}} \end{aligned}$

A particle is projected from the surface of the earth with a velocity equal to twice the escape velocity.…

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